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1 How many days would be holidays if we respect the birthdays of all the past emperors? Hajime Nanjo, Nobuhiro Shimizu, Taku Yamanaka Kuno and Yamanaka Groups End-of-the Year Presentation

2 Introduction 125 emperors in Japan in the past Birthdays of 3 emperors are holidays Meiji Emperor : Nov. 3, Culture Day Showa Emperor : Apr. 29, Showa Day Current Emperor : Dec. 23, The Emperor s Birthday 2

3 Question If we make the birthdays of all the past 125 emperors as holidays, how many Emperor s birthday holidays will we have? N = 365 (#days/year) M = 125 (#emperors) n : #holidays 3

4 1. Brute-force method Split emperors into multiple days Example: 7 emperors... Day A Day A B Day A B (7) (1+6) (2+5) 1 day 2 days 24 days Day A B C D ( ) 4 days

5 Example: #cases for the splitting pattern 365C1 : pick 1 day Day A B C D ( ) 4 days 7C1 : pick 1 emperor 365-1C3 : pick 3 days 7-1C2x3 : pick 2x3 emperors (2x3)! : Line up emperors 1/(2!) 3 : Remove double-counts = 1,833,153,121,800 5

6 #holidays partitions (7) (1+6) (1+1+5) ( ) ( ) ( ) ( ) (2+5) (1+2+4) ( ) ( ) (3+4) (1+3+3) ( ) (2+2+3) Partition Number ( ) = 15 #cases E E E E E+17 1 / Probability 1 #holidays 6 7

7 Problem of Partition No Partition Number N 100 It is LARGE! Took 100 min for M=90. Expect 80 hours for M=125 7

8 2. Monte Carlo Method Monte Carlo method can be used to calculate the expected number of holidays Jan. 1st M Dec. 31st M emperors are uniformly assigned to 365 boxes. hol hol hol hol ß Number of holidays is immediately obtained. 8

9 N histories

$of fraction of birth-month.$ $fraction The assignment of birth days is weighted according to this fraction.$

10 More realistic distribution Japanese government provides the statistic information of fraction of birth-month. fraction The assignment of birth days is weighted according to this fraction The real distribution is NOT uniform at all! Jan Sep Dec

11 N histories Expected number of holidays decreases by only 0.03 days (43 minutes). à Relax! Still we can take sufficient number of holidays even in this realistic situation.

12 3. Recurrence Formula Q(M n) : The number of cases for distributing M emperors to n specific days all in 1 day all in 2 days Q(M n) =n M nc 1 Q(M 1) nc 2 Q(M 2) min(m,n 1) = n M X i=1 Probability that M emperors fit in n days out of N days in a year P (M,N n) = N C n Q(M n) N M Used Python (infinite #digits for integers!) 12 nc i Q(M i)

13 Result of recurrence formula 0.1 Probability M = 125 emperors <n> = days RMS = 3.5 days n (#holidays) 13

14 Result of recurrence formula Probability n (#holidays) 14

15 4. Simple calculation N(365) patterns/row P workday =(1 1/N ) M P holiday =1 (1 1/N ) M 15

16 Result of simple calculation Binomial distribution : N,Pholiday P(n) Mean = N x Pholiday = 365 x (1-(1-1/365) 125 )=106 days RMS = NPholiday(1-Pholiday) = 8.7 days 16

17 <n> Comparisons Recurrence formula and the simple formula agree up to 12 digits. But why X n n N C n Q(M n)/n M = N 1 (1 1/N ) M? MC is consistent with the recurrence formula 300 <n> formula MC 0.04 MC-formula 0 M

18 RMS Comparisons RMS of the simple formula does not agree 8 formula MC + simple RMS M 18

19 #holidays = Conclusion N 1 (1 1/N ) M ± 3.5 days (for 125 emperors) Probability distribution function = Q(M n) P (M,N n) = N C n N M Q(M n) = n M min(m,n 1) i=1 nc i Q(M i) RMS needs the recurrence formula or MC 19

20 Conclusion Depending on how you approach/think, the problem can be extremely difficult and time consuming, or extremely simple and quick Monte Carlo is easy and robust Writeup is available on the program page 20

21 Happy Holidays! 21

22 Backup 22

23 Check with data nese Translate s page in: English Turn off for: Japanese Options Web member registration AKB48 Group ID Schedule Goods Ticket AKB48 Theater Discography Official mobile Password Fan club Chinese Handshake event Web member registration Select Language Japanese AKB48 Group ID Blog Password members Narrow your search er name A K B Theater 4 入山杏奈大家志津香小嶋菜月佐々木優佳里 Anna Iriyama Shizuka Oya Natsuki Kojima Yukari Sasaki Discography AKB48 Team A AKB48 Team A AKB48 Team A AKB48 Team A 8 mobile club Handshake ev MOfficial = 138 members Fan n = 114 diﬀerent birthdays expected n = ± 3.7 生 of birth ツゴー研究生公演ゴー研究生公演ツゴー研究生公演ゴー研究生公演白間美瑠田北香世子谷口めぐ中西智代梨 Miru Shiroma Kayoko Takita Megu Taniguchi Chiyori Nakanishi AKB48 Team A / NMB48 Team M AKB48 Team A AKB48 Team A AKB48 Team A 17年12月24日(日) y for tickets 樋渡結依宮崎美穂宮脇咲良横山由依 Yui Hiwatashi Miho Miyazaki Sakura Miyawaki Yui Yokoyama AKB48 Team A AKB48 Team A AKB48 Team A / HKT48 Team K IV AKB48 Team A 市川愛美久保怜音兒玉遥篠崎彩奈 Manami Ichikawa Satone Kubo Haruka Kodama Ayana Shinozaki AKB48 Team K AKB48 Team K AKB48 Team K / HKT48 Team H AKB48 Team K 23

24 Check with data n (days) sumo, basketball, boxing, volleyball politicians wrestling actors/actresses movie directors baseball players medical scholars tennis players jurists translators boat racers jockeys golf players Takarazuka AKB NMB48, SKE48 HelloPro HKT48 Nogizaka M

25 Differences between data and calculations n (data - expected) RMS 10 2 M

26 5. Discussion Expected number of holidays [ ( n = N ) ] M N [ ( = N ) N M ] N N N [ 1 exp ( M N )] (if N>>1) Why exponential?? 26

27 Actually, For 1 specific day, the expected number of emperors whose birthday is that day µ = M/N(= 125/365) Probability that some emperor s birthday is on that day, with Poisson distribution 1 P (µ, 0) = 1 e µ µ 0 0! =1 e µ Expected number of birthdays in a year with N days N 1 e M/N 27

28 How to count birthdays Example: Download to 076_1.txt, 076_2,txt, <a href=" </a> ( ) <img src="../images/death.gif" title=" "/> <a href=" </a> ( ) <img src="../images/ death.gif" title=" "/> cat 076_*.txt tr ' ' '\n' (split to lines) <a href=" </a> ( ) <img src="../images/death.gif" title=" "/> <a href=" </a> ( ) <img src="../images/death.gif" title=" "/> cat 076_*.txt tr ' ' '\n' grep 14px sed 's/\<a href.*//' sed s/.*\>//' grep sed 's/.* //'

29 cat 076_*.txt tr ' ' '\n' grep 14px sed 's/\<a href.*//' sed s/.*\>//' grep sed s/.* //' wc -l Count the number of lines 482 cat 076_*.txt tr ' ' '\n' grep 14px sed 's/\<a href.*//' sed s/.*\>//' grep sed s/.* //' sort -u wc -l Count the number of lines after removing duplicate lines

N C 1 7 C 1 : 1 1 : 1 1 N 1 C 3 : (2+2+2) 3 6 C (2 3)!/(2!) ,833,153,121, N M ( ) #holidays partitions

v1.0 2017 12 28 1 125 (11 3 ) (4 29 ) (12 23 ) 125 125 1 N = 365 M = 125 P (M, N n) n 2 7 (1) 7 1 (1+6) 1 1 6 1 (2+5) 2 1 5 1 (1+2+2+2) 1 1 2 1 2 1 2 1 15 (1+2+2+2) 4 1 1 N C 1 7 C 1 : 1 1 : 1 1 N 1 C